Question

$\Delta ABC$ and $\Delta DBC$ lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR || BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD.

Answer 1

Given: ΔABC and ΔDBC lie on the same side of BC. P is a point on BC, PQ || AB and PR || BD are drawn meeting AC at Q and CD at R respectively.

To Prove: QR || AD

Proof: In ΔABC

PQ || AB

⇒CP/PB=CQ/QA—(1) (by thales theorem)

In ΔBCD, PR || BD

Therefore,
CP/PB=CR/RD — (2) (by thales theorem)

From (1) & (2), we get:

CQ/QA=CR/RD

Hence, in ΔACD, Q and R the points in AC and CD such that

Therefore,
CQ/QA=CR/RD

QR || AD (by the converse of Thales theorem)

Hence proved